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A closed subset of a Lindelöf space is Lindelöf.

Let $M\subseteq X$ be closed and $\left\{U_t\right\}_{t\in T}$ be a cover of $M$; that is,

$$M\subseteq\bigcup_{t\in T}U_t.$$

By definition of the relative topology, for every $t\in T$ there exists a $V_t\subseteq X$ for which $U_t = M\cap V_t$.

Notice $U_t = M\cap V_t\subseteq V_t$, so that

$$M\subseteq\bigcup_{t\in T}U_t\subseteq\bigcup_{t\in T}V_t.$$

Therefore, $\left\{V_t\right\}_{t\in T}$ is a cover of $M$.

As $\left\{X\setminus M\right\}$ is a cover of $X\setminus M$, the collection of sets $\left\{X\setminus M\right\}\cup\left\{V_t\right\}_{t\in T}$ is a cover of $X$. Since $X$ is Lindelöf, there exists a countable set $S\subseteq T$ so that $\left\{X\setminus M\right\}\cup\left\{V_s\right\}_{s\in S}$ covers $X$; that is,

$$X\subseteq\bigcup_{s\in S}(X\setminus M)\cup(V_s).$$

However, from here the proof goes on to say that $\left\{M\cap V_s\right\}_{s\in S}$ covers $M$...how?

Rough_Manifolds
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3 Answers3

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Simple: if $x \in M$, then $$x \in X \subseteq (X \setminus M) \cup \bigcup_{s \in S} V_s.$$ Since $x \notin X \setminus M$, we must have $$x \in \bigcup_{s \in S} V_s.$$ Thus $x \in V_{s'}$ for some $s' \in S$. Since $x \in M$ as well, we get $$x \in V_{s'} \cap M \subseteq \bigcup_{s \in S} (V_s \cap M).$$ Hence, $$M \subseteq \bigcup_{s \in S} (V_s \cap M).$$

Theo Bendit
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Well, if $x \in M$ it is certainly not covered by the set $X\setminus M$ in the countable subcover $\{V_s\}_{s \in S} \cup \{X\setminus M\}$ for $X$, so it must be covered by one of the $V_s$, $s \in S$. So $x \in M \cap V_s$ for that $s$.

And this set equals $U_s$ by construction. So, in fact, any arbitrary $x \in M$ is covered by some $U_s$ for $s \in S$. Which finishes the proof that $M$ is Lindelöf.

Henno Brandsma
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The fact that $\{X\setminus M\}\cup\{V_s:s\in S\}$ covers $X$ means that

$$(X\setminus M)\cup\bigcup_{s\in S}V_s=X\,.$$

Of course this means that

$$M\subseteq (X\setminus M)\cup\bigcup_{s\in S}V_s\,,$$

since $M\subseteq X$. And $X\setminus M$ is disjoint from $M$, so it must be the case that

$$M\subseteq\bigcup_{s\in S}V_s\,,$$

i.e., that $\{V_s:s\in S\}$ covers $M$.

Brian M. Scott
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  • Understandable. However, idid you mean to write $\supseteq$? Or is it really an $=$ for the collection being an open cover? – Rough_Manifolds Mar 28 '21 at 09:26
  • @D.Math: You mean in the first displayed line? $\supseteq$ is enough for it to cover $X$, but in this case we know that all of the sets involved are subsets of $X$, so their union cannot include more than $X$: everything is happening in $X$. Thus, the union is in fact equal to $X$. – Brian M. Scott Mar 28 '21 at 18:20